Numerical Methods in Engineering with Python Phần 5 pps

raw_input"\nPress return to exit"Because the roots are computed with finite accuracy, each deflation introducessmall errors in the coefficients of the deflated polynomial.. First Noncent

165 4.6 Systems of Equations

The trajectory of a satellite orbiting the earth is

1+ e sin(θ + α) where (R, θ) are the polar coordinates of the satellite, and C, e, and α are con-

stants (e is known as the eccentricity of the orbit) If the satellite was observed at

the three positions

v

θ

A projectile is launched at O with the velocity v at the angle θ to the horizontal.

The parametric equation of the trajectory is

166 Roots of Equations

The three angles shown in the figure of the four-bar linkage are related by

150 cosθ1+ 180 cos θ2− 200 cos θ3 = 200

150 sinθ1+ 180 sin θ2− 200 sin θ3 = 0Determineθ1andθ2whenθ3= 75◦ Note that there are two solutions.

where T is the horizontal component of the cable force (it is the same in all

seg-ments of the cable) In addition, there are two geometric constraints imposed bythe positions of the supports:

−4 sin θ1− 6 sin θ2+ 5 sin θ2= −3

4 cosθ1+ 6 cos θ2+ 5 cos θ3= 12Determine the anglesθ1,θ2, andθ3

The polynomial equation P n (x) = 0 has exactly n roots, which may be real or

complex If the coefficients are real, the complex roots always occur in conjugate

pairs (x + ix , x − ix ), where x and x are the real and imaginary parts, respectively

167 ∗ 4.7 Zeroes of Polynomials

For real coefficients, the number of real roots can be estimated from the rule of

Descartes:

• The number of positive, real roots equals the number of sign changes in the

ex-pression for P n (x), or less by an even number.

• The number of negative, real roots is equal to the number of sign changes in

P n(−x), or less by an even number

As an example, consider P3(x) = x3− 2x2− 8x + 27 Because the sign changes twice, P3(x)= 0 has either two or zero positive real roots On the other hand,

P3(−x) = −x3− 2x2+ 8x + 27 contains a single sign change; hence P3(x) possesses

one negative real zero

The real zeroes of polynomials with real coefficients can always be computed byone of the methods already described But if complex roots are to be computed, it

is best to use a method that specializes in polynomials Here we present a methoddue to Laguerre, which is reliable and simple to implement Before proceeding to La-guerre’s method, we must first develop two numerical tools that are needed in anymethod capable of determining the zeroes of a polynomial The first of these is anefficient algorithm for evaluating a polynomial and its derivatives The second algo-

rithm we need is for the deflation of a polynomial, that is, for dividing the P n (x) by

x − r , where r is a root of P n (x) = 0.

Evaluation Polynomials

It is tempting to evaluate the polynomial in Eq (4.9) from left to right by the following

algorithm (we assume that the coefficients are stored in the array a):

p = 0.0 for i in range(n+1):

p = p + a[i]*x**i

Because x k is evaluated as x × x × · · · × x (k − 1 multiplications), we deduce that

the number of multiplications in this algorithm is

1+ 2 + 3 + · · · + n − 1 = 1

2n(n− 1)

If n is large, the number of multiplications can be reduced considerably if we evaluate

the polynomial from right to left For an example, take

P4(x) = a0+ a1x + a2x2+ a3x3+ a4x4

After rewriting the polynomial as

P (x) = a + x {a + x [a + x (a + xa )]}

P0(x) = a n

P i (x) = a n −i + x P i−1(x), i = 1, 2, , n (4.10)leading to the algorithm

proce-Some root-finding algorithms, including Laguerre’s method, also require

evalua-tion of the first and second derivatives of P n (x) From Eq (4.10) we obtain by

## module evalPoly

’’’ p,dp,ddp = evalPoly(a,x).

Evaluates the polynomial

p = a[0] + a[1]*x + a[2]*xˆ2 + + a[n]*xˆn with its derivatives dp = p’ and ddp = p’’

169 ∗ 4.7 Zeroes of Polynomials

dp = 0.0 + 0.0j ddp = 0.0 + 0.0j for i in range(1,n+1):

cause the degree of the polynomial is reduced every time a root is found Moreover,

by eliminating the roots that have already been found, the chances of computing thesame root more than once are eliminated

Laquerre’s formulas are not easily derived for a general polynomial P n (x) However,

the derivation is greatly simplified if we consider the special case where the

polyno-mial has a zero at x = r and (n − 1) zeroes at x = q Hence, the polynomial can be

170 Roots of Equations

written as

P n (x) = (x − r )(x − q) n−1 (a)Our problem is now this: given the polynomial in Eq (a) in the form

P n (x) = a0+ a1x + a2x2+ · · · + a n x n

determine r (note that q is also unknown) It turns out that the result, which is

ex-act for the special case considered here, works well as an iterative formula with anypolynomial

Differentiating Eq (a) with respect to x, we get

P n(x) = (x − q) n−1+ (n − 1)(x − r )(x − q) n−2

= P n (x)

$1

1 Let x be a guess for the root of P n (x)= 0 (any value will do)

2 Evaluate P n (x), P n(x), and P n(x) using the procedure outlined in Eqs (4.11).

3 Compute G(x) and H(x) from Eqs (4.14).

4 Determine the improved root r from Eq (4.16) choosing the sign that results in the

larger magnitude of the denominator (this can be shown to improve convergence).

5 Let x ← r and repeat steps 2–5 until |P n (x)| < ε or |x − r | < ε, where ε is the error

tolerance

171 ∗ 4.7 Zeroes of Polynomials

One nice property of Laguerre’s method is that it converges to a root, with very

few exceptions, from any starting value of x

polyRootsThe functionpolyRootsin this module computes all the roots of P n (x)= 0, where

the polynomial P n (x) defined by its coefficient array a = [a0, a1, , a n] After the firstroot is computed by the nested functionlaguerre, the polynomial is deflated using deflPolyand the next zero computed by applyinglaguerreto the deflated poly-

nomial This process is repeated until all n roots have been found If a computed root

has a very small imaginary part, it is more than likely that it represents roundoff error.Therefore,polyRootsreplaces a tiny imaginary part by zero

from evalPoly import * from numpy import zeros,complex from cmath import sqrt

from random import random def polyRoots(a,tol=1.0e-12):

def laguerre(a,tol):

x = random() # Starting value (random number)

n = len(a) - 1 for i in range(30):

for i in range(n-2,-1,-1):

b[i] = a[i+1] + root*b[i+1]

return b

n = len(a) - 1 roots = zeros((n),dtype=complex)

raw_input("\nPress return to exit")

Because the roots are computed with finite accuracy, each deflation introducessmall errors in the coefficients of the deflated polynomial The accumulated roundofferror increases with the degree of the polynomial and can become severe if the poly-nomial is ill conditioned (small changes in the coefficients produce large changes inthe roots) Hence, the results should be viewed with caution when dealing with poly-nomials of high degree

The errors caused by deflation can be reduced by recomputing each root usingthe original, undeflated polynomial The roots obtained previously in conjunctionwith deflation are employed as the starting values

P3(x) = 3x3+ 8x2− 2

EXAMPLE 4.11

A root of the equation P3(x) = x3− 4.0x2− 4.48x + 26.1 is approximately x = 3 − i.

Find a more accurate value of this root by one application of Laguerre’s iterativeformula

Solution Use the given estimate of the root as the starting value Thus,

x = 3 − i x2= 8 − 6i x3= 18 − 26i

Thanks to the good starting value, this approximation is already quite close to the

exact value r = 3.20 − 0.80i.

174 Roots of Equations

EXAMPLE 4.12

UsepolyRootsto compute all the roots of x4− 5x3− 9x2+ 155x − 250 = 0.

Solution The commands

>>> from polyRoots import *

>>> print polyRoots([-250.0,155.0,-9.0,-5.0,1.0])resulted in the output

Problems 6–9 A zero x = r of P n (x) is given Determine all the other zeroes of P n (x)

by using a calculator You should need no tools other than deflation and the quadraticformula

The most prominent root-finding algorithm omitted from this chapter is Brent’s

method, which combines bisection and quadratic interpolation It is potentially more

efficient than Ridder’s method, requiring only one function evaluation per iteration(as compared to two evaluations in Ridder’s method), but this advantage is somewhatnegated by elaborate bookkeeping

176 Roots of Equations

There are many methods for finding zeroes of polynomials Of these, the

Jenkins-Traub algorithm2deserves special mention because of its robustness and widespreaduse in packaged software

The zeroes of a polynomial can also be obtained by calculating the eigenvalues

of the n × n “companion matrix”

which is equivalent to P n (x) = 0 Thus the eigenvalues of A are the zeroes of P n (x).

The eigenvalue method is robust, but considerably slower than Laguerre’s method.But it is worthy of consideration if a good program for eigenvalue problems isavailable

2 M Jenkins and J Traub, SIAM Journal on Numerical Analysis, Vol 7 (1970), p 545.

5 Numerical Differentiation

Given the function f (x), compute d n f /dx n at given x

5.1 Introduction

Numerical differentiation deals with the following problem: We are given the

func-tion y = f (x) and wish to obtain one of its derivatives at the point x = x k The term

“given” means that we either have an algorithm for computing the function or

pos-sess a set of discrete data points (x i , y i ), i = 0, 1, , n In either case, we have cess to a finite number of (x, y) data pairs from which to compute the derivative If

ac-you suspect by now that numerical differentiation is related to interpolation, ac-you areright – one means of finding the derivative is to approximate the function locally by

a polynomial and then differentiate it An equally effective tool is the Taylor series

expansion of f (x) about the point x k, which has the advantage of providing us withinformation about the error involved in the approximation

Numerical differentiation is not a particularly accurate process It suffers from aconflict between roundoff errors (due to limited machine precision) and errors inher-ent in interpolation For this reason, a derivative of a function can never be computedwith the same precision as the function itself

5.2 Finite Difference Approximations

The derivation of the finite difference approximations for the derivatives of f (x) is based on forward and backward Taylor series expansions of f (x) about x, such as

f (x + h) = f (x) + hf(x)+h2

2!f

(x)+h33!f

(x)+h44!f(4)(x)+ · · · (a)

f (x − h) = f (x) − hf(x)+h2

2!f

(x)−h33!f

(x)+h44!f(4)(x)− · · · (b)

177

We also record the sums and differences of the series:

f (x + h) + f (x − h) = 2f (x) + h2f(x)+ h4

12f(4)(x)+ · · · (e)

can be solved for various derivatives of f (x) The number of equations involved and

the number of terms kept in each equation depend on the order of the derivative andthe desired degree of accuracy

First Central Difference Approximations

The solution of Eq (f ) for f(x) is

f(x)= f (x + h) − f (x − h)

6 f

(x)− · · ·or

f(x)= f (x + h) − f (x − h)

which is called the first central difference approximation for f(x) The term O(h2)

reminds us that the truncation error behaves as h2

Similarly, Eq (e) yields the first central difference approximation for f(x):

f(x)= f (x + h) − 2f (x) + f (x − h)

12f(4)(x) +

or

f(x)= f (x + h) − 2f (x) + f (x − h)

Central difference approximations for other derivatives can be obtained from

Eqs (a)–(h) in the same manner For example, eliminating f(x) from Eqs (f ) and (h) and solving for f(x) yields

f(x)= f (x + 2h) − 2f (x + h) + 2f (x − h) − f (x − 2h)

179 5.2 Finite Difference Approximations

f(4)(x)= f (x + 2h) − 4f (x + h) + 6f (x) − 4f (x − h) + f (x − 2h)

is available from Eqs (e) and (g) after eliminating f(x) Table 5.1 summarizes the

results

First Noncentral Finite Difference Approximations

Central finite difference approximations are not always usable For example,

con-sider the situation where the function is given at the n discrete points x0, x1, , x n

Because central differences use values of the function on each side of x, we would

be unable to compute the derivatives at x0and x n Clearly, there is a need for finite

difference expressions that require evaluations of the function on only one side of x These expressions are called forward and backward finite difference approximations.

Noncentral finite differences can also be obtained from Eqs (a)–(h) Solving Eq

(a) for f(x), we get

Keeping only the first term on the right-hand side leads to the first forward difference

Second Noncentral Finite Difference Approximations

Finite difference approximations ofO(h) are not popular, for reasons to be explained

shortly The common practice is to use expressions ofO(h2) To obtain noncentraldifference formulas of this order, we have to retain more terms in the Taylor series As

an illustration, we derive the expression for f(x) We start with Eqs (a) and (c), which

f (x + 2h) = f (x) + 2hf(x) + 2h2f(x)+4h3

3 f

(x)+2h4

3 f(4)(x)+ · · ·

We eliminate f(x) by multiplying the first equation by 4 and subtracting it from the

second equation The result is

f (x + 2h) − 4f (x + h) = −3f (x) − 2hf(x)+h4

2 f(4)(x)+ · · ·

Therefore,

f(x)=−f (x + 2h) + 4f (x + h) − 3f (x)

4 f(4)(x)+ · · ·

or

f(x)= −f (x + 2h) + 4f (x + h) − 3f (x)

Equation (5.8) is called the second forward finite difference approximation.

The derivation of finite difference approximations for higher derivatives involve

additional Taylor series Thus the forward difference approximation for f(x) utilizes

181 5.2 Finite Difference Approximations

Table 5.3b Coefficients of backward finite difference approximations ofO(h2)

series for f (x + h), f (x + 2h), and f (x + 3h); the approximation for f(x) involves Taylor expansions for f (x + h), f (x + 2h), f (x + 3h), f (x + 4h), and so on As you can

see, the computations for high-order derivatives can become rather tedious The sults for both the forward and backward finite differences are summarized in Tables5.3a and 5.3b

re-Errors in Finite Difference Approximations

Observe that in all finite difference expressions, the sum of the coefficients is zero

The effect on the roundoff error can be profound If h is very small, the values of

f (x), f (x ± h), f (x ± 2h), and so forth will be approximately equal When they are

multiplied by the coefficients and added, several significant figures can be lost On

the other hand, we cannot make h too large, because then the truncation error would

become excessive This unfortunate situation has no remedy, but we can obtain somerelief by taking the following precautions:

• Use double-precision arithmetic.

• Employ finite difference formulas that are accurate to at least O(h2)

To illustrate the errors, let us compute the second derivative of f (x) = e −x at

x= 1 from the central difference formula, Eq (5.2) We carry out the calculations

with six- and eight-digit precision, using different values of h The results, shown in Table 5.4, should be compared with f(1)= e−1= 0.367 879 44.

In the six-digit computations, the optimal value of h is 0 08, yielding a result

ac-curate to three significant figures Hence, three significant figures are lost because of a

combination of truncation and roundoff errors Above optimal h, the dominant error

is due to truncation; below it, the roundoff error becomes pronounced The best sult obtained with the eight-digit computation is accurate to four significant figures

differ-Because the extra precision decreases the roundoff error, the optimal h is smaller

(about 0.02) than in the six-figure calculations.

5.3 Richardson Extrapolation

Richardson extrapolation is a simple method for boosting the accuracy of certain merical procedures, including finite difference approximations (we also use it later inother applications)

nu-Suppose that we have an approximate means of computing some quantity G Moreover, assume that the result depends on a parameter h Denoting the approxi- mation by g(h), we have G = g(h) + E(h), where E(h) represents the error Richard- son extrapolation can remove the error, provided that it has the form E (h) = ch p , c and p being constants We start by computing g(h) with some value of h, say, h = h1

In that case we have

which is the Richardson extrapolation formula It is common practice to use h2=

h1/2, in which case Eq (5.8) becomes

G= 2p g(h1/2) − g(h1)

183 5.3 Richardson Extrapolation

Let us illustrate Richardson extrapolation by applying it to the finite difference

approximation of (e −x)at x= 1 We work with six-digit precision and utilize the sults in Table 5.4 Because the extrapolation works only on the truncation error, we

re-must confine h to values that produce negligible roundoff In Table 5.4 we have

g(0.64) = 0.380 610 g(0.32) = 0.371 035

The truncation error in central difference approximation is E (h) = O(h2)= c1h2+

c2h4+ c3h6+ Therefore, we can eliminate the first (dominant) error term if we substitute p = 2 and h1= 0.64 in Eq (5.9) The result is

Use the data in Example 5.1 to compute f(0) as accurately as you can

Solution One solution is to apply Richardson extrapolation to finite difference

ap-proximations We start with two forward difference approximations ofO(h2) for f(0):

Recall that the error in both approximations is of the form E (h) = c1h2+ c2h4+

c3h6+ We can now use Richardson extrapolation, Eq (5.9), to eliminate the inant error term With p= 2 we obtain

d − a cos α − b cos β2

+a sin α + b sin β2

− c2= 0For a given value ofα, we can solve this transcendental equation for β by one of the

root-finding methods in Chapter 2 This was done withα = 0◦, 5◦, 10◦, , 30◦, theresults being

If link A B rotates with the constant angular velocity of 25 rad/s, use finite difference

approximations ofO(h2) to tabulate the angular velocity d β/dt of link BC against α.

Solution The angular speed of BC is

185 5.4 Derivatives by Interpolation

where d β/dα can be computed from finite difference approximations using the data

in the table Forward and backward differences ofO(h2) are used at the endpoints,central differences elsewhere Note that the increment ofα is

by the derivative of the interpolant This method is particularly useful if the data

points are located at uneven intervals of x, when the finite difference approximations

listed in the last section are not applicable.1

Polynomial Interpolant

The idea here is simple: fit the polynomial of degree n

P n−1(x) = a0+ a1x + a2x2+ · · · + a n x n

through n + 1 data points and then evaluate its derivatives at the given x As pointed

out in Section 3.2, it is generally advisable to limit the degree of the polynomial toless than 6 in order to avoid spurious oscillations of the interpolant Because theseoscillations are magnified with each differentiation, their effect can devastating Inview of this limitation, the interpolation is usually a local one, involving no more than

a few nearest-neighbor data points

For evenly spaced data points, polynomial interpolation and finite differenceapproximations produce identical results In fact, the finite difference formulas areequivalent to polynomial interpolation

1 It is possible to derive finite difference approximations for unevenly spaced data, but they would not be as accurate as the formulas derived in Section 5.2.

186 Numerical Differentiation

Several methods of polynomial interpolation were introduced in Section 3.2 fortunately, none of them are suited for the computation of derivatives of the inter-

Un-polant The method that we need is one that determines the coefficients a0, a1, , a n

of the polynomial There is only one such method discussed in Chapter 3: the

least-squares fit Although this method is designed mainly for smoothing of data, it will

carry out interpolation if we use m = n in Eq (3.22) – recall that m is the degree of the interpolating polynomial and n+ 1 represents the number of data points to be fitted

If the data contains noise, then the least-squares fit should be used in the smoothing

mode, that is, with m < n After the coefficients of the polynomial have been found,

the polynomial and its first two derivatives can be evaluated efficiently by the tionevalPolylisted in Section 4.7

func-Cubic Spline Interpolant

Because of to its stiffness, the cubic spline is a good global interpolant; moreover, it

is easy to differentiate The first step is to determine the second derivatives k iof thespline at the knots by solving Eqs (3.11) This can be done with the functioncurva- turesin the modulecubicSplinelisted in Section 3.3 The first and second deriva-tives are then computed from

&

3(x − x i)2

x i − x i+1 − (x i − x i+1)

'+y i − y i+1

nearest-Solution of Part (1) The interpolant is P2(x) = a0+ a1x + a2x2passing through the

points at x = 1.9, 2.1, and 2.4 The normal equations, Eqs (3.22), of the least-squares